Free Math Worksheets

- Which operation shows how many times a smaller price fits into a larger price? - Use a related multiplication fact or break apart the larger number.

1. Divide the bicycle price by the light price: \(360 \div 6 = 60\). 2. Therefore, the bicycle costs \(60\) times as much as the light.

The bicycle costs \(60\) times as much as the light.

- Divide each greater height by the small tree's height. - You can divide both numbers by \(10\) before completing each division.

1. Divide the large tree's height by the small tree's height: \(320 \div 40 = 8\). 2. Divide the pole's height by the small tree's height: \(800 \div 40 = 20\).

a) The large tree is \(8\) times as tall. b) The utility pole is \(20\) times as tall.

- Ask about the missing number of tickets. - “Four times as many” indicates multiplication. - Break \(62\) into tens and ones if helpful.

1. One possible question is, “How many tickets were sold on Saturday?” 2. Multiply: \(62 \times 4 = 248\).

Question: “How many tickets were sold on Saturday?” Answer: The theater sold \(248\) tickets on Saturday.

- Ask about the missing price. - Write a question that begins with “How much...?” - “Three times as much” indicates multiplication.

1. One possible question is, “How much does the helmet cost?” 2. Multiply: \(\$14 \times 3 = \$42\).

Question: “How much does the helmet cost?” Answer: The helmet costs \(\$42\).

- Convert both measurements in a pair to the same unit. - Recall the conversion factors for yards, feet, inches, pounds, ounces, and tons. - Divide the larger numerical value by the smaller one.

1. a) Since \(1\,\text{yd}=3\,\text{ft}\), the first measurement is \(3\) times as large. 2. b) Since \(1\,\text{ft}=12\,\text{in.}\), the first measurement is \(12\) times as large. 3. c) Since \(1\,\text{lb}=16\,\text{oz}\), the first measurement is \(16\) times as large. 4. d) Since \(1\,\text{ton}=2000\,\text{lb}\), calculate \(2000 \div 200=10\). The first measurement is \(10\) times as large.

a) \(3\) times b) \(12\) times c) \(16\) times d) \(10\) times

- Think about how many bucketfuls fit into the aquarium. - Which operation separates a total into equal-size groups? - Use a related multiplication fact to check the division.

1. Divide the aquarium's capacity by the bucket's capacity: \(200\,\text{L} \div 5\,\text{L} = 40\). Therefore, \(40\) full buckets are needed. 2. Since \(40\) bucketfuls fit in the aquarium, the aquarium holds \(40\) times as much water as the bucket.

a) \(40\) full buckets b) \(40\) times as much water

- Look for the question that asks how many times one amount fits into another. - Use a related multiplication fact to calculate \(360 \div 9\).

1. Question C asks how many times as much, so it is the multiplicative comparison. 2. Divide: \(360 \div 9 = 40\).

C. The tank holds \(40\) times as much water as the watering can.

- Compare \(60\) with \(20\) and determine how many groups of \(20\) are in \(60\). - Write a multiplication equation and its related division equation.

1. There are three groups of \(20\) in \(60\), so \(3 \times 20 = 60\). 2. The related division equation is \(60 \div 20 = 3\).

There are three times as many tomato plants as pepper plants: \(3 \times 20 = 60\) and \(60 \div 20 = 3\).

- Find Sarah’s score first. - Use Sarah’s score to find Tim’s score. - “Twice as many” means multiply by \(2\).

1. Sarah’s score is twice Lucas’s score: \(135{,}000\times 2=270{,}000\). 2. Tim’s score is twice Sarah’s score: \(270{,}000\times 2=540{,}000\).

Sarah scored \(270{,}000\) points. Tim scored \(540{,}000\) points.

- Ask how many times the smaller weight fits into the greater weight. - Use a question that includes “times as heavy.” - Find a related multiplication fact.

1. One possible question is, “How many times as heavy is the large dog as the small dog?” 2. Divide the greater weight by the smaller weight: \(42 \div 6 = 7\).

Question: “How many times as heavy is the large dog as the small dog?” Answer: The large dog is \(7\) times as heavy as the small dog.

- Find the price of one flower in each package. - Then compare the two unit prices using division. - Check whether the larger unit price is a whole-number multiple of the smaller one.

1. One rose costs \(\$32 \div 8 = \$4\). 2. One tulip costs \(\$10 \div 5 = \$2\). 3. Since \(\$4 \div \$2 = 2\), one rose costs twice as much as one tulip.

One rose costs \(2\) times as much as one tulip.

- How many hours are in a full day and night? - Subtract the nighttime hours to find the remaining daylight hours. - Which operation tells how many times one amount fits into another?

1. Find the daylight period: \(24\,\text{hr} - 18\,\text{hr} = 6\,\text{hr}\). 2. Compare the durations: \(18\,\text{hr} \div 6\,\text{hr} = 3\). The night is \(3\) times as long as the daylight period.

There are \(6\) hours of daylight. The night is \(3\) times as long as the daylight period.

- What operation finds half of an amount? - First find the number of daylight hours in December. - Subtract the daylight hours from the full \(24\)-hour day.

1. Find the December daylight period: \(12\,\text{hr} \div 2 = 6\,\text{hr}\). 2. Find the nighttime period: \(24\,\text{hr} - 6\,\text{hr} = 18\,\text{hr}\).

The night lasts \(18\) hours.

- First find the cost of one crate of pansies. - How can you find how many times the smaller price fits into the larger price? - Which operation compares prices multiplicatively?

1. Find the cost of one crate of pansies: \(\$20 \div 4 = \$5\). 2. Compare the prices by division: \(\$15 \div \$5 = 3\).

A crate of roses costs \(3\) times as much as a crate of pansies.

- Can you compare the group's \(48\) muffins directly with the teacher's \(24\)? - What must you know about one child to make a fair comparison? - Imagine each child's muffins shown separately beside the teacher's muffins.

1. Find the number of muffins baked by one child: \(48 \div 6 = 8\). 2. Compare the teacher's amount with one child's amount: \(24 \div 8 = 3\). 3. The total of \(48\) muffins represents six children together, so it cannot be compared directly with the work of one teacher. A per-person amount is needed for a fair comparison.

a) The teacher bakes \(3\) times as many muffins as one child. b) You must first find one child's amount so that the comparison is between two individual people.

- First find the number of pages in the reference section. - Once you know both page counts, determine how many times the smaller number fits into the larger one. - Removing a factor of ten from both numbers may make the division easier.

1. Find the number of pages in the reference section: \(240 - 210 = 30\). 2. Compare the page counts by division: \(210 \div 30 = 7\). The main text has seven times as many pages as the reference section.

The main text has \(7\) times as many pages as the reference section.

- How much does the paint set cost without the paintbrush? - Compare the cost of the paint set with the cost of the paintbrush. - Which operation finds how many times one amount fits into another?

1. Find the cost of the paint set: \(\$45 - \$5 = \$40\). 2. Compare the costs by division: \(\$40 \div \$5 = 8\). The paint set costs eight times as much as the paintbrush.

The paint set costs \(8\) times as much as the paintbrush.

- First find the total number of yellow tulips. - Compare the number of yellow tulips with the number of red tulips. - How many times does the number of red tulips fit into the number of yellow tulips?

1. Find the number of yellow tulips: \(9 + 18 = 27\). 2. Compare the numbers by division: \(27 \div 9 = 3\). There are three times as many yellow tulips as red tulips.

There are \(3\) times as many yellow tulips as red tulips.

- First find Sophie’s exact score. - Use place value to simplify the multiplicative comparison. - How many copies of Lucas’s score make Sophie’s score?

1. Find Sophie's score: \(150 + 450 = 600\). 2. Compare the scores by division: \(600 \div 150 = 4\). Sophie scores four times as many points as Lucas.

Sophie scores \(4\) times as many points as Lucas.

- Use each multiplicative comparison to find the amount raised by each class. - Add the three amounts. - Compare the total with the equipment cost.

1. Class 4B raises \(3 \times \$135 = \$405\). 2. Class 4C raises \(2 \times \$405 = \$810\). 3. Together, the classes raise \(\$135 + \$405 + \$810 = \$1350\). 4. Since \(\$1350 < \$1500\), they did not raise enough. 5. They are short by \(\$1500 - \$1350 = \$150\).

No. The classes raised \(\$1350\), so they are \(\$150\) short.

- First find the total after the number of apples becomes five times as large. - The question asks only for the added apples. - Subtract the original amount from the new total.

1. Find the new total: \(24 \times 5 = 120\). 2. Subtract the original apples: \(120 - 24 = 96\).

The vendor adds \(96\) apples.

- First find the current number of streetlights. - The question asks for the increase, not the new total. - Subtract the old number from the new number.

1. Find the current number: \(8 \times 6 = 48\). 2. Find the increase: \(48 - 8 = 40\).

The number of streetlights increased by \(40\).

- First find the amount of orange juice. - “Four times as many” indicates multiplication. - Then add both amounts.

1. Find the amount of orange juice: \(15\,\text{L} \times 4 = 60\,\text{L}\). 2. Add both kinds of juice: \(15\,\text{L} + 60\,\text{L} = 75\,\text{L}\).

There are \(75\,\text{L}\) of juice altogether.

- First find the total cost of all the hot dog packages. - Compare that total with the amount spent on mustard. - Which operation finds how many times one amount fits into another?

1. Find the total cost of the hot dogs: \(15 \times \$4 = \$60\). 2. Compare the hot dog cost with the mustard cost: \(\$60 \div \$6 = 10\). Mr. Weber spends ten times as much on hot dogs as on mustard.

He spends \(10\) times as much on hot dogs as on mustard.

- How much money remains after the soccer balls are purchased? - The remaining money is spent on jump ropes. - Compare the soccer ball cost with the remaining amount. - How many times does the smaller amount fit into the larger amount?

1. Find the amount spent on jump ropes: \(\$100 - \$80 = \$20\). 2. Compare the two amounts: \(\$80 \div \$20 = 4\). The class spends four times as much on soccer balls as on jump ropes.

The class spends \(4\) times as much on soccer balls as on jump ropes.

- First find the second-year harvest. - The question asks for the difference, not the new total. - Subtract the first-year harvest from the second-year harvest.

1. Find the second-year harvest: \(15\,\text{kg} \times 6 = 90\,\text{kg}\). 2. Find the increase: \(90\,\text{kg} - 15\,\text{kg} = 75\,\text{kg}\).

The second-year harvest is \(75\,\text{kg}\) greater.

- First find the cost of one chair. - Then find the cost of one table. - How many times does the smaller unit price fit into the larger one?

1. Find the cost of one chair: \(\$400 \div 8 = \$50\). 2. Find the cost of one table: \(\$300 \div 2 = \$150\). 3. Compare the unit prices: \(\$150 \div \$50 = 3\).

One table costs \(3\) times as much as one chair.

- First find the cost of one basketball. - Then find the cost of one medicine ball. - Check whether the medicine-ball price is exactly ten times the basketball price.

1. Find the cost of one basketball: \(\$54 \div 6 = \$9\). 2. Find the cost of one medicine ball: \(\$270 \div 3 = \$90\). 3. Compare the prices: \(\$90 \div \$9 = 10\). The teacher is correct.

Yes. One basketball costs \(\$9\), and one medicine ball costs \(\$90\). Since \(90\) is ten times \(9\), the teacher is correct.

- Ask about the unknown amount from last year. - The earlier amount must be smaller because this year's amount is three times as much. - Division reverses multiplication by \(3\). - Break \(135\) into parts that are easy to divide by \(3\).

1. One possible question is, “How many kilograms of honey did the beekeeper harvest last year?” 2. Since this year's amount is three times last year's amount, divide: \(135 \div 3 = 45\,\text{kg}\).

Question: “How many kilograms of honey did the beekeeper harvest last year?” Answer: The beekeeper harvested \(45\,\text{kg}\) last year.

- “Four times as many” indicates multiplication. - Once you know both amounts, subtract to find how many more. - Break apart \(32\) if helpful.

1. Find the number of green apples: \(32 \times 4 = 128\). 2. Find the difference: \(128 - 32 = 96\).

There are \(128\) green apples, which is \(96\) more than the number of red apples.

- First find the total number of storybooks. - Once you know both numbers, determine how many times the smaller fits into the larger. - What operation does “more than” suggest for the first step?

1. Find the number of storybooks: \(15 + 45 = 60\). 2. Compare the numbers by division: \(60 \div 15 = 4\). There are four times as many storybooks as nonfiction books.

There are \(4\) times as many storybooks as nonfiction books.

- How many liters does the large tank hold altogether? - How many groups of \(25\,\text{L}\) make the large tank's capacity? - Counting by twenty-fives may help.

1. Find the capacity of the large tank: \(25\,\text{L} + 175\,\text{L} = 200\,\text{L}\). 2. Compare the capacities by division: \(200 \div 25 = 8\). The large tank holds eight times as much water as the small tank.

The large tank holds \(8\) times as much water as the small tank.

- First find the elephant's current age. - Once you know both ages, find their difference. - Does the age difference between two living things change over time?

1. Find the elephant's current age: \(60 \div 5 = 12\) years. 2. Find the difference in their ages: \(60 - 12 = 48\) years. 3. Their age difference stays constant, so the turtle was \(48\) years old when the elephant was born.

The turtle was \(48\) years old when the elephant was born.

- First find the total number of chairs in each section. - Division compares two quantities when the question asks “how many times as many.” - How many times does the smaller total fit into the larger total?

1. Find the number of chairs in the front section: \(4 \times 18 = 72\). 2. Find the number of chairs in the back section: \(3 \times 8 = 24\). 3. Compare the totals by division: \(72 \div 24 = 3\).

The front section has \(3\) times as many chairs as the back section.

- First find the total number of each kind of roll. - Which operation compares quantities when the question asks “how many times as many”? - Divide the larger total by the smaller total.

1. Find the number of plain rolls: \(5 \times 36 = 180\). 2. Find the number of multigrain rolls: \(3 \times 15 = 45\). 3. Compare the totals by division: \(180 \div 45 = 4\).

The baker makes \(4\) times as many plain rolls as multigrain rolls.

- Find Lucas’s and Sophie’s numbers separately. - “Less than” indicates subtraction. - To determine how many times as large one number is, divide the larger number by the smaller number.

1. Lucas’s number is \(48 \div 4 = 12\). 2. Sophie’s number is \(100 - 4 = 96\). 3. Divide to make the multiplicative comparison: \(96 \div 12 = 8\). 4. Therefore, Sophie’s number is \(8\) times as large as Lucas’s number.

Sophie’s number is \(8\) times as large as Lucas’s number.

- Replace each adult ticket with two equal child-ticket shares. - Count the total number of child-ticket shares. - Divide the total cost by the number of shares, then double the child price.

1. One adult ticket costs the same as two child tickets. Therefore, \(3\) adult tickets have the same cost as \(3\times2=6\) child tickets. 2. Together with the \(4\) actual child tickets, the total cost is equivalent to \(6+4=10\) child tickets. 3. A child ticket costs \(\$70\div10=\$7\). 4. An adult ticket costs \(2\times\$7=\$14\).

Child ticket: \(\$7\) Adult ticket: \(\$14\)

- Rewrite the entire fountain-pen weight in eraser units. - Determine how many eraser units equal the fountain pen. - Use the relationship between a pencil package and an eraser.

1. Two packages of pencils weigh as much as \(2\times4=8\) erasers. 2. Including the two additional erasers, the fountain pen weighs as much as \(8+2=10\) erasers. 3. One eraser weighs \(200\,\text{g}\div10=20\,\text{g}\). 4. One package of pencils weighs \(4\times20\,\text{g}=80\,\text{g}\).

Eraser: \(20\,\text{g}\) Package of pencils: \(80\,\text{g}\)

- Represent each child’s amount with equal-sized parts. - Find the total number of parts. - Divide \(120\) by the total number of parts. - Check that the three amounts add to \(120\).

1. Represent Ava’s amount as \(1\) equal part. Ben receives \(2\) parts, and Chloe receives \(1+2=3\) parts. 2. Altogether, there are \(1+2+3=6\) equal parts. 3. Each part is \(120\div 6=20\) marbles. 4. Ava receives \(20\), Ben receives \(2\times 20=40\), and Chloe receives \(3\times 20=60\).

Ava receives \(20\) marbles, Ben receives \(40\), and Chloe receives \(60\).

- Represent the geese as one part and the ducks as four equal parts. - Add the new geese to find the total in part b. - For part c, compare the unchanged number of ducks with the new number of geese.

1. Geese represent \(1\) equal part and ducks represent \(4\) parts, for \(5\) parts altogether. 2. The number of geese is \(150\div 5=30\), and the number of ducks is \(4\times 30=120\). 3. After \(10\) geese arrive, there are \(150+10=160\) birds. 4. There are now \(30+10=40\) geese. Since \(120\div 40=3\), there are only \(3\) times as many ducks as geese.

a) \(120\) ducks and \(30\) geese b) \(160\) birds c) No. There are now \(40\) geese, and \(120\div 40=3\).

- First find the number of red packages. - “Four times as many” indicates multiplication. - Add the blue and red amounts.

1. Find the number of red packages: \(135 \times 4 = 540\). 2. Add both colors: \(135 + 540 = 675\).

The store has \(675\) packages altogether.

- First find the number of relay-race awards. - Add the awards for both events. - Compare the total needed with \(1000\).

1. Find the number of relay-race awards: \(124 \times 7 = 868\). 2. Find the total needed: \(124 + 868 = 992\). 3. Since \(992 < 1000\), there are enough awards.

Yes. The school needs \(992\) awards, so the \(1000\) printed awards are enough.

- First convert both measurements to the same smaller unit. - Divide to determine how many equal parts fit into the whole. - For numbers ending in zeros, use place value to simplify the division.

1. a) Convert \(1\,\text{m}\) to \(100\,\text{cm}\). Then \(100 \div 20=5\), so \(20\,\text{cm}\) fits \(5\) times. 2. b) Convert \(1\,\text{kg}\) to \(1000\,\text{g}\). Then \(1000 \div 250=4\), so \(250\,\text{g}\) fits \(4\) times. 3. c) Convert \(1\,\text{km}\) to \(1000\,\text{m}\). Then \(1000 \div 50=20\), so \(50\,\text{m}\) fits \(20\) times.

a) \(5\) times b) \(4\) times c) \(20\) times

- The “four times” statement describes the two-week total. - Subtract the first-week pages from that total. - Compare the second-week amount with \(3 \times 112\).

1. Find the total after two weeks: \(112 \times 4 = 448\). 2. Find the second-week amount: \(448 - 112 = 336\). 3. Check the claim: \(112 \times 3 = 336\), so the student is correct.

a) The class reads \(336\) pages during the second week. b) Yes. Since \(112 \times 3 = 336\), the second-week amount is three times the first-week amount.

- First find Finn's total number of cards. - “Four times as many” indicates multiplication. - Subtract Lucas's amount to find how many more.

1. Find Finn's number of cards: \(48 \times 4 = 192\). 2. Find the difference: \(192 - 48 = 144\).

Finn has \(144\) more cards than Lucas.

- First find the number of cups of water. - “Three times as many” indicates multiplication. - Subtract the apple-juice amount from the water amount.

1. Find the number of cups of water: \(115 \times 3 = 345\). 2. Find the difference: \(345 - 115 = 230\).

The fair prepares \(230\) more cups of water than apple juice.

- First find the number of mystery books. - “Three times as many” indicates multiplication. - Subtract the nonfiction amount from the mystery amount.

1. Find the number of mystery books: \(45 \times 3 = 135\). 2. Find the difference: \(135 - 45 = 90\).

There are \(90\) more mystery books than nonfiction books.

- First find Marie's total savings. - The question asks for the difference, not Marie's total. - Subtract Lucas's amount from Marie's amount.

1. Find Marie's savings: \(\$160 \times 4 = \$640\). 2. Find the difference: \(\$640 - \$160 = \$480\).

Marie has saved \(\$480\) more than Lucas.

- Represent the two prices as equal parts. - Count the total number of parts. - Divide the total cost by the number of parts.

1. The pump represents \(1\) equal part, and the soccer ball represents \(6\) equal parts. 2. The total is \(1 + 6 = 7\) equal parts. 3. One part is \(\$42 \div 7 = \$6\). 4. The pump costs \(\$6\), and the soccer ball costs \(6 \times \$6 = \$36\).

The pump costs \(\$6\), and the soccer ball costs \(\$36\).

- Use equal parts to represent “three times as many.” - Removing \(4\) fruits and adding \(4\) others keeps the total unchanged. - Compare the two new amounts at the end.

1. Pears represent \(1\) equal part and apples represent \(3\) parts, for \(4\) parts altogether. 2. Each part is \(48\div 4=12\), so there are \(12\) pears and \(3\times 12=36\) apples. 3. After the change, there are \(36-4=32\) apples and \(12+4=16\) pears. 4. The difference is \(32-16=16\).

a) \(36\) apples and \(12\) pears b) There are \(16\) more apples than pears.

- First find the total number of computers in the labs. - Include the tablets in today’s total. - Divide today’s total by the original number of devices.

1. The computer labs contain \(3\times 28=84\) computers. 2. The school now has \(84+42=126\) digital devices. 3. Compare with the original amount: \(126\div 14=9\). 4. The school now owns \(9\) times as many devices.

The school owns \(9\) times as many digital devices today.

- Represent the youngest person's age with one equal part. - How many of those parts represent each of the other ages? - How many equal parts make the total of \(66\) years?

1. Let Leo's age be represented by \(1\) equal part. 2. Maya's age is \(2\) parts. 3. Their father's age is \(4 \times 2=8\) parts. 4. Altogether, their ages make \(1+2+8=11\) equal parts. 5. One part is \(66\div11=6\) years. 6. Their father is \(8 \times 6=48\) years old.

Their father is \(48\) years old.

- Express the two longer pieces in terms of the shortest piece. - How many equal parts make the full \(135\,\text{cm}\) length? - After finding the shortest length, use the comparisons to find the other lengths.

1. Represent the shortest piece with \(1\) equal part. 2. The middle piece is \(2\) parts. 3. The longest piece is \(3 \times 2=6\) parts. 4. The entire rope is \(1+2+6=9\) equal parts. 5. One part is \(135\div9=15\,\text{cm}\). 6. The three lengths are \(15\,\text{cm}\), \(2 \times 15=30\,\text{cm}\), and \(6 \times 15=90\,\text{cm}\).

The shortest piece is \(15\,\text{cm}\), the middle piece is \(30\,\text{cm}\), and the longest piece is \(90\,\text{cm}\).

- First find the July harvest. - “Six times as large” indicates multiplication. - Subtract the June harvest to test the claim.

1. Find the July harvest: \(12\,\text{kg} \times 6 = 72\,\text{kg}\). 2. Find the difference: \(72\,\text{kg} - 12\,\text{kg} = 60\,\text{kg}\). 3. The statement is correct.

Yes. The July harvest is \(72\,\text{kg}\), which is exactly \(60\,\text{kg}\) more than the June harvest.

- First find the number needed for the park. - “Eight times as many” indicates multiplication. - Add both amounts.

1. Find the number needed for the park: \(24 \times 8 = 192\). 2. Add the amounts for the bed and park: \(192 + 24 = 216\).

The nursery needs \(216\) geraniums altogether.

- Use the employee's comparison to find the number of tickets still available. - Once you know the sold and unsold amounts, how can you find the total? - Breaking \(480\) into hundreds and tens may help with the division.

1. Find the number of tickets still available: \(480 \div 4 = 120\). 2. Add the tickets already sold and the tickets still available: \(480 + 120 = 600\).

There were \(600\) tickets available for the concert altogether.

- Express both prices in the same unit. - One dollar equals \(100\) cents. - Think about what happens to the comparison factor when the smaller price is cut in half.

1. Convert the art set price: \(\$8 = 800\) cents. 2. For part a), \(800 \div 40 = 20\). 3. For part b), \(800 \div 20 = 40\).

a) The art set costs \(20\) times as much. b) The art set costs \(40\) times as much.

- Recall the conversion from liters to milliliters. - For part b), find how many groups of \(25\,\text{g}\) make \(1000\,\text{g}\). - If the comparison unit doubles in length, what happens to the number of times it fits into the same whole?

1. a) Since \(1\,\text{L}=1000\,\text{mL}\), calculate \(1000 \div 200=5\). 2. b) Since \(1\,\text{kg}=1000\,\text{g}\), calculate \(1000 \div 25=40\). 3. c) Since \(1\,\text{m}=100\,\text{cm}\), calculate \(100\,\text{cm} \div 2\,\text{cm}=50\). A \(2\,\text{cm}\) segment is twice as long as a \(1\,\text{cm}\) segment, so it fits half as many times.

a) \(5\) portions b) \(40\) objects c) \(50\) times; doubling the smaller segment halves the number of segments that fit.

- When the red gummy bears are removed from the total, the yellow gummy bears remain. - Use the statement that there are three times as many red gummy bears as yellow gummy bears. - Add the two amounts to find the total.

1. The difference between the total number and the number of red gummy bears is the number of yellow gummy bears. Therefore, there are \(12\) yellow gummy bears. 2. There are three times as many red gummy bears: \(3\times12=36\). 3. The total is \(12+36=48\).

There are \(48\) gummy bears in the bowl.

- Subtract the cost of the guides first. - Rewrite each adult ticket as two child-ticket shares. - Divide the remaining ticket cost by the total number of shares.

1. The three guides cost \(3\times\$8=\$24\). 2. The tickets cost \(\$80-\$24=\$56\) altogether. 3. One adult ticket costs the same as two child tickets. The \(3\) adult tickets are equivalent to \(6\) child-ticket shares. Including the \(2\) child tickets gives \(8\) shares. 4. A child ticket costs \(\$56\div8=\$7\), and an adult ticket costs \(2\times\$7=\$14\).

Adult ticket: \(\$14\) Child ticket: \(\$7\)

- Represent each person’s amount with equal shares. - In part b, count Lucas as two shares. - In part c, first split the prize into two equal parts.

1. a) Divide equally: \(\$1800\div4=\$450\) per person. 2. b) Lucas receives \(2\) shares, and each other friend receives \(1\) share, for \(2+1+1+1=5\) shares. One share is \(\$1800\div5=\$360\). Lucas receives \(2\times\$360=\$720\), and each other friend receives \(\$360\). 3. c) Lucas and the other three friends together receive equal halves of the prize. Lucas receives \(\$1800\div2=\$900\). The other \(\$900\) is divided by \(3\), so each friend receives \(\$300\).

a) Each person receives \(\$450\). b) Lucas receives \(\$720\); each other friend receives \(\$360\). c) Lucas receives \(\$900\); each other friend receives \(\$300\).

- Represent the amounts with equal shares. - In part b, begin with one share for chess. - Count the total shares before finding the value of one share. - Add the three amounts to check the total.

1. a) Divide equally among three programs: \(\$2100\div3=\$700\). 2. b) Let the chess program receive \(1\) share. Then tennis receives \(2\) shares, and soccer receives \(4\) shares. There are \(1+2+4=7\) shares altogether. 3. One share is \(\$2100\div7=\$300\). 4. Chess receives \(\$300\), tennis receives \(2\times\$300=\$600\), and soccer receives \(4\times\$300=\$1200\).

a) Each program receives \(\$700\). b) Chess: \(\$300\); tennis: \(\$600\); soccer: \(\$1200\)

- Remove Class A’s extra \(40\) seedlings first. - The amount left represents two equal copies of the combined total for Classes B and C. - Divide the combined amount for Classes B and C into a \(3\)-to-\(1\) share model.

1. Let the combined number planted by Classes B and C be one amount. Class A plants that same amount plus \(40\). 2. Remove the extra \(40\) from the total: \(120-40=80\). The remaining \(80\) represents two equal combined amounts, so Classes B and C together plant \(80\div2=40\) seedlings. 3. Class A plants \(40+40=80\) seedlings. 4. Class B plants three times as many as Class C, so their combined \(40\) seedlings are divided into \(3+1=4\) equal shares. One share is \(40\div4=10\). 5. Class C plants \(10\) seedlings, and Class B plants \(3\times10=30\) seedlings.

Class A: \(80\) seedlings Class B: \(30\) seedlings Class C: \(10\) seedlings